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561hw1sol

# 561hw1sol - Math 561 H Homework 1 Solutions Fall 2011 Drew...

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Math 561 H Fall 2011 Homework 1 Solutions Drew Armstrong 1. Let G be a group. Prove that the identity element of G is unique (that is, there is only one element of G satisfying the defining property of an identity element). This justifies our use of the special symbol “ e ”. Proof. Suppose that there exist e, f G such that ae = ea = a for all a G and bf = fb = b for all b G . Letting a = f and b = e yields e = ef = f . 2. Let G be a finite group. (a) Show that there are an odd number of x G such that x 3 = e . (b) Show that there are an even number of x G such that x 2 6 = e . (Hint: What is the inverse of x n ?) Proof. Given x G we first note that ( x n ) - 1 = ( x - 1 ) n (make sure you know why this is true). To prove part (a) we want to show that the size of the set S = { x G : x 3 = e } is odd. First note that we have x S x - 1 S because inverting both sides of x 3 = e gives ( x - 1 ) 3 = e - 1 = e . Then substituting x x - 1 in the same argument gives x - 1 S x S . Thus the elements of S come in pairs ( x, x - 1 ) with x 6 = x - 1 and possibly some single elements with x = x - 1 . But x S and x = x - 1 together imply that e = x 2 x = x 3 x = e . Hence e S is the only element of S that does not come from a pair ( x, x - 1 ). We conclude that | S | is odd. To prove part (b) let T = { x G : x 2 6 = e } . As above we can see that x T x - 1 T . Furthermore, by definition no element of T satisfies x = x - 1 . Hence the elements of T come in pairs ( x, x - 1 ) and we conclude that | T | is even. 3. Let G be a finite group. For all a, b G , show that ab and ba have the same order as elements of G . Proof. We wish to show that the sets h ab i = { ( ab ) k : k Z } and h ba i = { ( ba ) k : k Z } have the same size. This will be done if we can find a bijection from h ba i to h ab i . We claim that φ a from Problem 4 is such a bijection.

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561hw1sol - Math 561 H Homework 1 Solutions Fall 2011 Drew...

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